\begin{answer}
    We have

$$
\begin{aligned}
\frac{\partial }{\partial^2 \eta}\int
p(y;\eta)dy &= \frac{\partial}{\partial \eta}\int\frac{\partial}{\partial \eta }p(y;\eta)dy\\
&= \frac{\partial}{\partial \eta}\int (y - a'(\eta))b(y) \exp(\eta y - a(\eta))dy\\
&= \frac{\partial}{\partial \eta}\int yb(y)  \exp(\eta y - a(\eta))dy - a''(\eta)\\
&= \int (y^2 - a'(\eta)) \exp(\eta y - a(\eta))dy = a''(\eta)\\
&= E[y^2;\eta] - E[y;\eta]^2 - a''(\eta)\\
&= Var(y;\eta) - a''(\eta) = 0
\end{aligned}
$$

And thus

$$
Var(y;\eta) = a''(\eta)
$$

\end{answer}
